Chapter 14, Problem 81A

Interpretation Introduction

Interpretation:

Required amount of Na₂CO₃is to be calculated.

Concept introduction:

Molality: Molality is defined as the number of moles of solute per kg of the solvent.

In expression,

$\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofthesolute}{Molecularweightofthesolute}$

Answer to Problem 81A

The mass of Na₂CO₃ is 134.6 gm.

Explanation of Solution

Given information:

Molality of Na₂CO₃ solution = 8.20 mol/kg

Mass of the solvent water = 155 gm = 0.155 kg

Write the expression for Molality.

$\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

Calculate the number of moles of $N{a}_{2}C{O}_3$.

$\begin{array}{l}MolalityofN{a}_{2}C{O}_{3}solution=\frac{NumberofmolesofN{a}_{2}C{O}_3}{ Mass of water}\\ NumberofmolesofN{a}_{2}C{O}_3=MolalityofN{a}_{2}C{O}_{3}solution\times Mass of water\\ =8.20mol/kg\times 0.155kg=1.27mole\end{array}$

Calculate the mass of $N{a}_{2}C{O}_3$.

$\begin{array}{l}NumberofmolesofN{a}_{2}C{O}_3=\frac{MassofN{a}_{2}C{O}_3}{MolecularweightofN{a}_{2}C{O}_3}\\ MassofN{a}_{2}C{O}_3=NumberofmolesofN{a}_{2}C{O}_3\times MolecularweightofN{a}_{2}C{O}_3\\ =1.27mol\times 106gm/mol=134.6gm\end{array}$

(Note: Molecular weight of $N{a}_{2}C{O}_3$ = 106 gm/mol)

So, the mass of Na₂CO₃ is 134.6 gm.